In this lesson, we will learn how
to solve problems about the equilibrium of a particle under the action of three
forces meeting at a point using resultant of forces or triangle of forces
Remember, when two or more forces
act on a rigid body and that body does not accelerate in any direction, the forces
are in equilibrium. That doesn’t necessarily mean that
the object is at rest. It could also be moving at a
Take, for instance, two forces of
equal magnitude acting on a body in opposite directions but along the same line of
action. These two forces are in
equilibrium, and the body will either remain at rest or continue to move at a
constant velocity. Suppose we have a different pair of
forces acting on a rigid body. For that body to be in equilibrium,
there must also be a third force acting on the body which is equal in magnitude and
opposite in direction to the resultant of those two forces.
Now, if we know the angle 𝜃
between 𝐴 and 𝐵, then we can calculate the magnitude of the resultant using this
formula. Let’s apply this to a very quick
example. Suppose the magnitude of 𝐴 is five
newtons and the magnitude of 𝐵 is four newtons. If the body is in equilibrium, then
we can calculate the magnitude of the force 𝐅 by finding the magnitude of the
resultant of 𝐴 and 𝐵. That’s the square root of five
squared plus four squared plus two times five times four times cos of 70
degrees. That’s the square root of 47.84 and
so on, which is roughly 6.916. This means that the magnitude of
the force 𝐅 is 6.92 newtons to two decimal places.
Let’s increase the difficulty
during our first full example.
Use the following diagram to
find the tension in 𝐶𝐵. Round your answer to two
Remember, if a pair of forces
are acting on a rigid body and that body is in equilibrium, then a third force
𝐅 must also be acting on the body, which is equal in magnitude and opposite in
direction to the resultant of those two forces. In this case, the pair of
forces are the tensions with magnitudes 𝑇 sub one and 𝑇 sub two. The third force is the downward
force of 10 newtons, meaning that the magnitude of the resultant 𝑅 must also be
So if we know the angle between
the two forces, let’s call that 𝜃, we can use this equation to find the
magnitude of the resultant. Substituting what we know about
this system and we get 10 equals the square root of 𝑇 sub one squared plus 𝑇
sub two squared plus two times 𝑇 sub one times 𝑇 sub two cos 𝜃.
Since in this case the triangle
is equilateral, 𝑇 sub one and 𝑇 sub two are equal. So we can replace these with
𝑇, square both sides, and begin to evaluate the right. We can now factor two 𝑇
squared, so 100 equals two 𝑇 squared times one plus cos 𝜃. Making 𝑇 squared the subject
by dividing through by two times one plus cos 𝜃 gives 𝑇 squared equals 50 over
one plus cos 𝜃.
Next, we might spot that we can
find the value of 𝜃 using the length of the sides in the triangle. We label our triangle as
shown. And we get 50 squared equals 30
squared plus 30 squared minus two times 30 times 30 cos 𝜃. That’s 2500 equals 1800 minus
1800 cos 𝜃, and we can rearrange to find that cos of 𝜃 equals negative seven
over 18. Let’s substitute that into our
earlier expression. That’s 𝑇 squared equals 50
over one plus negative seven over 18, which is 900 over 11.
Finally, we know that the
tension in 𝐶𝐵 is 𝑇 sub one, which is equal to 𝑇, so we just need to square
root. The square root of 900 over 11
is 9.0453. Correct to two decimal places,
that’s 9.05, so the tension in 𝐶𝐵 is 9.05 newtons.
When a rigid body is in equilibrium
under the action of three coplanar forces meeting at a point, we can analyze the
system using a triangle of forces. Let’s think about the forces as
vectors and represent them using arrows whose lengths are proportional to their
magnitude. Since the system is in equilibrium,
the sum of these forces is zero. Then, the addition of forces can be
represented by placing the arrows head to tail, as shown here. Let’s define this all formally. The force vectors that form a
triangle where the directions of the forces are all clockwise around the triangle or
counterclockwise have a zero resultant so are in equilibrium. Let’s demonstrate this in our next
Three coplanar forces 𝐅 sub
one, 𝐅 sub two, and 𝐅 sub three are acting on a body in equilibrium. The triangle of forces forms a
right triangle as shown. Given that the magnitude of 𝐅
sub one equals five newtons and the magnitude of 𝐅 sub two equals 13 newtons,
find the magnitude of 𝐅 sub three.
Since the body is in
equilibrium, we know that the vector sum of the three forces must be equal to
zero. We also know that this means we
can represent the system as a triangle, where the lengths of the triangle are
proportional to their magnitudes. This means we can treat the two
force magnitudes as if they were lengths and use the Pythagorean theorem to find
the missing magnitude. That is, five squared plus the
magnitude of 𝐅 sub three squared equals 13 squared. Subtracting five squared from
both sides and we see that the magnitude of 𝐅 sub three squared is equal to
Finally, we take the positive
square root. Remember, we don’t need a
negative root since a magnitude, by definition, is positive. The magnitude of this force is
12 newtons. So, we’ve found the magnitude
of 𝐅 sub three.
We were able to apply the
Pythagorean theorem here because the forces are proportional to the side lengths of
the triangle. This allows us to form an equation
which we define as the triangle of forces rule. The ratio of the magnitude of 𝐅
sub one and the length 𝐴𝐵 is equal to the ratio of the magnitude of 𝐅 sub two and
the length 𝐵𝐶, which in turn is equal to the ratio of the magnitude of 𝐅 sub
three and 𝐴𝐶.
In the figure, three forces of
magnitudes 𝐅 sub one, 𝐅 sub two, and 𝐅 sub three newtons meet at a point. The lines of action of the
forces are parallel to the sides of the right triangle. Given that the system is in
equilibrium, find the ratio of 𝐅 sub one to 𝐅 sub two to 𝐅 sub three.
Remember, when the three forces
are in equilibrium, the magnitudes of the forces are proportional to the side
lengths of the triangle. Let’s begin by finding the
length of the third side. We can label the vertices of
the triangle as shown, meaning we want to find the length of the hypotenuse,
𝐵𝐶. Let’s use the Pythagorean
theorem, so 𝐵𝐶 squared equals 87 squared plus 208.8 squared. That gives us 𝐵𝐶 squared
equals 51166.44. Taking the positive square root
gives us the length 𝐵𝐶 to be 226.2 centimeters.
Since the triangle is in
equilibrium, we know that the ratios of the forces and the side lengths they are
parallel to are all equal. We can therefore also say that
the ratio of two of the forces must be equal to the ratio of the respective side
lengths. Since 𝐴𝐵 is 87 centimeters
and 𝐴𝐶 is 208.8 centimeters, we can divide these to get five twelfths. In a similar way, we can find
the ratio of 𝐅 sub one and 𝐅 sub three by dividing 𝐴𝐵 by 𝐵𝐶. That gives us 𝐅 sub one over
𝐅 sub three equals five over 13. Since the numerator is the same
for each fraction, we can create the required ratio. The ratio of 𝐅 sub one to 𝐅
sub two to 𝐅 sub three is five to 12 to 13.
Of course, this process doesn’t
only work for right triangles. We can repeat a similar process
when working with non-right triangles. Let’s see what that looks like.
A body is under the effect of
three forces of magnitudes 𝐅 sub one, 𝐅 sub two, and 36 newtons, acting in the
directions 𝐴𝐵, 𝐵𝐶, and 𝐶𝐴, respectively, where triangle 𝐴𝐵𝐶 is a
triangle such that 𝐴𝐵 equals four centimeters, 𝐵𝐶 is six centimeters, and
𝐴𝐶 is six centimeters. Given that the system is in
equilibrium, find 𝐅 sub one and 𝐅 sub two.
Remember, for the system of
forces to be in equilibrium, there must be a zero resultant force. This means that the magnitude
of the force in our triangle must be in the same ratio as the length of the
sides of the triangle. So let’s sketch that out. We have an isosceles triangle,
with one side of length four centimeters and two of length six centimeters. We can trace the corresponding
force diagram over the top, where the ratio of 𝐅 sub one and the length of side
𝐴𝐵 is equal to the ratio of 𝐅 sub two and side 𝐵𝐶, which in turn is equal
to the ratio of the 36-newton force and the length of the third side.
If we write the ratios out for
the two sides of equal lengths, we should be able to spot the value of the
magnitude of 𝐅 sub two. The only way for the statement
to be true is if 𝐅 sub two is 36 newtons. Similarly, let’s compare the
first ratio with the one for side 𝐴𝐵. We can calculate the multiplier
for the side lengths by dividing four by six, to get two-thirds. So, we also need to multiply 36
by two-thirds to find the value of 𝐅 sub one. That’s 24 newtons. So, 𝐅 sub one is 24 newtons
and 𝐅 sub two is 36 newtons.
In our final example, let’s
look at how to apply this process to a system involving a suspended object.
A uniform rod of length 50
centimeters and weight 143 newtons is freely suspended at its end from the
ceiling by means of two perpendicular strings attached to the same point on the
ceiling. Given that the length of one of
the strings is 30 centimeters, determine the tension in each string.
Let’s begin by sketching this
system out. Here is the rod, supported by
two pieces of string that meet at an angle of 90 degrees. We might begin by calculating
the length of the third side in this triangle. Let’s call that 𝑙
centimeters. This will be useful. Since we know the system is in
equilibrium, so we will be able to then find the forces in the system. Since we have a right triangle,
we can use the Pythagorean theorem. 50 squared equals 30 squared
plus 𝑙 squared. Subtracting 30 from both sides
and we get 𝑙 squared equals 50 squared minus 30 squared, which is 1600. Finally, if we take the square
root of both sides of this equation, we find that 𝑙 is equal to 40.
Next, we know that the forces
acting here are the weights of the rod and the tensions in the strings. Since the forces are in
equilibrium, they can be drawn acting at the same point. The weight of the rod acts
vertically downward. Then, we can represent the
tension in the 40-centimeter string as 𝑇 sub one and the tension in the other
string as 𝑇 sub two. This gives us the corresponding
Finally, we know that the ratio
of the side lengths of the triangle and the corresponding forces are equal. That’s 143 over 50 equals 𝑇
sub one over 30. We can solve for 𝑇 sub one by
multiplying through by 30, which gives us 85.8 newtons.
Let’s repeat this for 𝑇 sub
two. This time ,143 over 50 equals
𝑇 sub two over 40. So we find 𝑇 sub two equals
143 over 50 times 40, which is 114.4 newtons. So the tensions in the two
pieces of string are 85.8 newtons and 114.4 newtons.
Let’s now recap the key points from
In this lesson, we saw that when a
rigid body is in equilibrium under the action of three coplanar forces that meet at
a point, we can analyze the system using a triangle of forces. And we learnt that in these cases,
the magnitude of the forces are proportional to the side lengths of the
triangle. This can allow us to calculate
missing forces and missing lengths.